Applying SFINAE pattern with universal reference

问题

I've got the following code:

#include <utility>
#include <iostream>
#include <queue>
#include <functional>
#include <stdio.h>
#include <future>

class Runner {
public:
    virtual void run() = 0;
    virtual ~Runner() {
    }
};

class Command: public Runner {
public:
    Command() {
        std::cout << "constructor" << std::endl;
    }
    virtual void run() {
    }
};

#define EXTEND(T, F) , typename = typename std::enable_if<std::is_base_of<F, typename std::decay<T>::type>::value, typename std::decay<T>::type>::type
#define NOT_EXTEND(T, F) , typename = typename std::enable_if<!std::is_base_of<F, typename std::decay<T>::type>::value, typename std::decay<T>::type>::type

class Executor {
private:
    std::queue<std::function<void(void)> > q;
public:
    template<class T EXTEND(T, Runner)>
    void push(T&& toBepushed) {
        q.push(
                std::bind(&std::decay<T>::type::run,
                        std::forward<T>(toBepushed)));
    }
    template<typename T NOT_EXTEND(T, Runner)>
    void push(T&& toBepushed) {
        q.push(std::forward<T>(toBepushed));
    }
    void perform() {
        std::function<void(void)>&& f = std::move(q.front());
        f();
    }
};

int main() {
    Executor b;
    Command c;
    b.push(c);
    b.perform();
    return 0;
}

When I compile I've got the following error:

g++ -std=c++0x -O2 -g3 -Wall -c -fmessage-length=0 -MMD -MP -MF"main.d" -MT"main.o" -o "main.o" "../main.cpp" ../main.cpp:45:7: error: ‘template void Bar::push(T&&)’ cannot be overloaded void push(T&& toBepushed) { ^~~~ ../main.cpp:39:7: error: with ‘template void Bar::push(T&&)’ void push(T&& toBepushed) { ^~~~ make: *** [subdir.mk:20: main.o] Error 1

I'm trying to apply SFINAE in order to have the push method applied according to the type used. How to solve?

回答1:

Default argument is not part of the template parameter list, thus your two push have identical template parameter list, which leads a redeclaration.

This is a typical drawback of typename = std::enable_if_t<...> approach. Instead, you should use std::enable_if_t<..., int> = 0 approach.

Change to this:

#define EXTEND(T, F) , typename std::enable_if<std::is_base_of<F, typename std::decay<T>::type>::value, int>::type = 0
#define NOT_EXTEND(T, F) , typename std::enable_if<!std::is_base_of<F, typename std::decay<T>::type>::value, int>::type = 0

来源：https://stackoverflow.com/questions/53351945/applying-sfinae-pattern-with-universal-reference

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c++

c++11

sfinae